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9 tháng 6 2016

1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2

= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)

= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)

= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)

= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)

= 1/90 - ( 1 - 1/9)

= 1/90 - 8/9

= 1/90 - 80/90

= -79/90

9 tháng 6 2016

1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2

= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)

= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)

= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)

= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)

= 1/90 - ( 1 - 1/9)

= 1/90 - 8/9

= 1/90 - 80/90

= -79/90

 mk nha cac ban 

\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{1}{90}-\left(\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\frac{8}{9}\)

\(=\frac{-79}{90}\)

25 tháng 8 2016

1/90 - 1/72 - 1/56 - ... - 1/6 - 1/2

= 1/90 - (1/2 + 1/6 + ... + 1/56 + 1/72)

= 1/90 - (1/1×2 + 1/2×3 + ... + 1/7×8 + 1/8×9)

= 1/90 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/7 - 1/8 + 1/8 - 1/9)

= 1/90 - (1 - 1/9)

= 1/90 - 8/9

= 1/90 - 80/90

= -79/90

25 tháng 8 2016

1/90 - 1/72 - 1/56 - ... - 1/6 - 1/2

= 1/90 - (1/2 + 1/6 + ... + 1/56 + 1/72)

= 1/90 - (1/1×2 + 1/2×3 + ... + 1/7×8 + 1/8×9)

= 1/90 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/7 - 1/8 + 1/8 - 1/9)

= 1/90 - (1 - 1/9)

= 1/90 - 8/9

= 1/90 - 80/90

= -79/90

25 tháng 8 2016

\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{1}{90}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=\frac{-79}{90}\)

8 tháng 9 2016

em lớp 6  nha

B= 1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42 + 1/56 + 1/72

B= 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9

B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

B=1+0-0-0-0-0-0-0-1/9

B=1-1/9

B=8/9

k em nha

8 tháng 9 2016

phần A em chịu

13 tháng 3 2017

    -1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2

= -1/2-1/6-1/12-1/20-1/30-1/42-1/56-1/64-1/72-1/90

= -(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/64+1/72+1/90)

= -(1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10)

= -(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)

= -(1-1/10)

= - 9/10

14 tháng 7 2021

Đây đề lớp 6 mà. Bn này làm đúng rồi nè!

11 tháng 9 2016

 Đặt biểu thức cần tính là A. Ta có : 
A = 9/10 -( 1/90 + 1/72 + ... + 1/2) 
= 9/10 - { 1/( 9.10) + 1/(9.8) + ... + 1/( 2.1)} 
= 9/10 - ( 1/9 - 1/10 + 1/8 - 1/9 + ...+ 1 - 1/2) ( 1/90 = 1/(9.10) = 1/9 - 1/10) 
= 9/10 - ( 1 - 1/10)  

12 tháng 9 2016

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

=\(1-\frac{1}{10}=\frac{9}{10}\)

k cho mk nha

12 tháng 9 2016

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(1-\frac{1}{10}\)

\(\frac{9}{10}\)

9 tháng 2 2021

Ta đặt A=\(-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(\Rightarrow A=-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)\(-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\) 

= - \(\left(1-\dfrac{1}{10}\right)=-\left(\dfrac{10-1}{10}\right)=-\dfrac{9}{10}\)

Ta có: \(-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)

\(=-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}\right)\)

\(=-\left(-\dfrac{1}{10}+1\right)\)

\(=-\left(1-\dfrac{1}{10}\right)\)

\(=-\left(\dfrac{10}{10}-\dfrac{1}{10}\right)=-\dfrac{9}{10}\)

15 tháng 3 2019

\(=\frac{9}{10.11}-\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)

\(=\frac{9}{10.11}-\frac{10-9}{9.10}-\frac{9-8}{8.9}-...-\frac{2-1}{1.2}\)

\(=\frac{9}{10.11}-\frac{10}{9.10}+\frac{9}{9.10}-...-\frac{2}{1.2}+\frac{1}{1.2}\)

\(=\frac{9}{10.11}-\frac{1}{9}+\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-\frac{1}{7}+\frac{1}{8}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)

\(=\frac{9}{10.11}+\frac{1}{10}-1\)

\(=-\frac{9}{11}\)